Advertisements
Advertisements
प्रश्न
An ideal gas is taken in a cyclic process as shown in the figure. Calculate
- work done by the gas
- work done on the gas
- Net work done in the process
Advertisements
उत्तर
a. Work done by the gas (along with AB)
W = P × ΔV = 600 × 3 = 1800 J
W = 1.8 kJ
b. Work is done on the gas (along with BC)
W = −P × ΔV = −400 × 3 = −1200 J
W = −1.2 kJ
c. Net work done in the process = Area under the curve AB
= Rectangle area + triangle area
= `("l" xx "b") + (1/2 xx "b" xx "h")`
= `(400 xx 3) + (1/2 xx 3 xx 200)`
= 1200 + 300
= 1500 J
W = 1.5 kJ
APPEARS IN
संबंधित प्रश्न
Write a note on free expansion.
When you exercise in the morning, by considering your body as a thermodynamic system, which of the following is true?
The V-T diagram of an ideal gas which goes through a reversible cycle A→B→C→D is shown below. (Processes D→A and B→C are adiabatic)
The corresponding PV diagram for the process is (all figures are schematic)
In an isochoric process, we have ____________.
Draw the PV diagram for the isothermal process.
Draw the PV diagram for the adiabatic process.
An ideal gas is made to go from a state A to stale B in the given two different ways (see figure) (i) an isobaric and then an isochoric process and (ii) an isochoric and then an isobaric process. The work done by gas in the two processes are W1 and W2 respectively. Then,
One mole of an ideal gas with `gamma` = 1.4 is adiabatically compressed so that its temperature rises from 27° C to 47° C. The change in the internal energy of the gas is (R = 8.3 J/mol.K) ____________.
Ideal gas for which 'ϒ' = 1.5 is suddenly compressed to `1/4`th of its initial volume. The ratio of 4 the final pressure to the initial pressure is ______.
`(ϒ = "C"_"p"/"C"_"v")`
In the figure shown here, the work done in the process ACBA is ______.
